Statistics Report: Analyzing Data with Histograms and Functions

Abdias Medina MECE 3360 1/27/2024 Statistics Report 1.) a.) After combining the voltage signal sets into one set, I calculated the standard deviation using the “STDEV” function, the max (Using MAX function), and min (Using MIN function). With these values I began calculating the bin sizes for each option using the provided equations. Once I had the bin sizes, I calculated the bins and used the data analysis tool to create the histograms for all four options. b.) I used the STDEV and AVERAGE functions to calculate the standard deviation and sample set mean which came out to be 0.580209682 and 0.000721508, respectively. 2.) a.) After inputting all 20,000 data, I halved the data four times and inputted it alongside the previous data point. Leaving 40 groups with 500 data points. b.) At the bottom of each column, I used the AVERAGE function to calculate the mean and the STDEV.S function to calculate the Standard deviation. I dragged these two equations across all columns to obtain my new data points. c.) I then inputted the new mean and standard deviation data points into a separate worksheet and calculated their means, standard deviation, minimum, maximum, and range using the previously mentioned functions. For the mean set, the results I obtained were: 0.000690506, 0.027813636, -0.04469, 0.065957, and 0.110642824, respectively. For the standard deviation set, the results I obtained were: 0.577594858, 0.013201596, 0.555316, 0.614369, and 0.059053301, respectively. d.) Using the needed values alongside the given option formulas, I calculated the bin sizes and bins for each option for the mean data sets. For the bin sizes I obtained 0. 0.028465, 0.017494, 0.020652, and 0.07087 for the four options. With the bin sizes I obtained the bins for each option and used the data analysis tool to graph their histogram using the results as outputs and the 40 data points as inputs. e.) Similarly, I used the values from the standard deviation data sets to calculate the bin sizes from each option formula and then the bins from the bin sizes. I in turn got 0.013511, 0.009337, 0.011022, and 0.553318 for the four option bin sizes. Using the bins as outputs and the 40 data sets as inputs, I created the histograms for each four options.

a. Option 1 has 27 bins of size 0.074813. The data appears uniformly distributed based on the plot. Figure-1.1 Option 1 Histogram Option 2 has 142 bins of size 0.01414. The data appears uniformly distributed based on the plot. Figure-1.2 Option 2 Histogram 0 200 400 600 800 1000 Frequency Bin Histogram Frequency 0 20 40 60 80 100 120 140 160 180 200 -1 -0.901017422 -0.802034843 -0.703052265 -0.604069686 -0.505087108 -0.40610453 -0.307121951 -0.208139373 -0.109156794 -0.010174216 0.088808363 0.187790941 0.286773519 0.385756098 0.484738676 0.583721255 0.682703833 0.781686411 0.88066899 0.979651568 Frequency Bin Histogram Frequency

Option 3 has 16 bins of size 0.139962. The data appears uniformly distributed. Figure-1.3 Option 3 Histogram Option 4 has 99 bins of size 0.020153. The data appear uniformly distributed. Figure 1.3 Option 4 Histogram b.) Mean = 0.00072 Standard Deviation = 0.5802 0 500 1000 1500 2000 -1 -0.860037778 -0.720075556 -0.580113334 -0.440151112 -0.30018889 -0.160226668 -0.020264445 0.119697777 0.259659999 0.399622221 0.539584443 0.679546665 0.819508887 0.959471109 1.099433331 More Frequency Bin Histogram Frequency 0 50 100 150 200 250 -1 -0.899236796 -0.798473593 -0.697710389 -0.596947186 -0.496183982 -0.395420779 -0.294657575 -0.193894371 -0.093131168 0.007632036 0.108395239 0.209158443 0.309921646 0.41068485 0.511448054 0.612211257 0.712974461 0.813737664 0.914500868 More Frequency Bin Histogram Frequency

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c.) Mean = 0.0006905. Standard Deviation = 0.02781. d.) Option 1 has 4 bins of size 0.028465. The data appear normally distributed. Figure 2.1 Mean Option 1 Histogram Option 2 has 7 bins of size 0.017495. The data appear normally distributed. Figure 2.2 Mean option 2 Histogram Option 3 has 6 bins of size 0.020652. The data appears normally distributed. 0 5 10 15 20 Frequency Bin Histogram Frequency 0 5 10 15 Frequency Bin Histogram Frequency

Figure 2.3 Mean option 3 Histogram Option 4 has 10 bins of size 0.012164. The graph appears normally distributed. Figure 2.4 Mean option 4 Histogram e.) Option 1 has 5 bins of size 0.013511. The data appears normally distributed. 0 5 10 15 Frequency Bin Histogram Frequency 0 2 4 6 8 10 Frequency Bin Histogram Frequency

Figure 3.1 Standard Deviation Option 1 Histogram Option 2 has 7 bins of size 0.009337. The data appear normally distributed. Figure 3.2 Standard Deviation Option 2 Histogram Option 3 has 6 bins of size 0.011022. The graph appears to be uniformly distribute but shifted to the left. 0 5 10 15 20 Frequency Bin Histogram Frequency 0 2 4 6 8 10 12 Frequency Bin Histogram Frequency

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Figure 3.3 Standard Deviation Option 3 Histogram Option 4 has 9 bins of size 0.006492. The graph appears uniformly distributed. Figure 3.4 Standard Deviation Option 4 Histogram 0 5 10 15 20 Frequency Bin Histogram Frequency 0 2 4 6 8 10 Frequency Bin Histogram Frequency

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